Question

The given function \[f(x)= \log (x+ \sqrt{x^2+1})\] is an even function ? odd function ? or a periodic function ?

Answer 1

do you know how to test if function is even odd?

Answer 2

Yeah
f(x)= f(-x) Even
f(x)=f(x) Odd

Answer 3

if you just put x=1 and x=-1 and check with calculator results are same just one positive and other negative so it's odd, but idk how to show with x :/

Answer 4

\[f(-x)=\log(-x+\sqrt{x^2+1})\]

Answer 5

@ishaan94 knows :D

Answer 6

Ishaan's not here !

Answer 7

you need to show that \[\log(-x+\sqrt{x^2+1})=-\log(x+\sqrt{x^2+1})\] but i don't know how :/

Answer 8

okay :)

Answer 9

None :/

Answer 10

any even function is that, where f(x)=f(-x), that is, basically, the function takes same values for any particular 'x' as well as '-x'
so, the function's graph is symmetrical about the Y-axis.
something like this... (considering any arbitrary function)



so you may have guessed what you have to do in here.. tell me if i need to explain more.